Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k + 2}{k - 10} \div \dfrac{k^2 - 4k - 12}{4k - 24} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k + 2}{k - 10} \times \dfrac{4k - 24}{k^2 - 4k - 12} $ First factor the quadratic. $q = \dfrac{k + 2}{k - 10} \times \dfrac{4k - 24}{(k + 2)(k - 6)} $ Then factor out any other terms. $q = \dfrac{k + 2}{k - 10} \times \dfrac{4(k - 6)}{(k + 2)(k - 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 2) \times 4(k - 6) } { (k - 10) \times (k + 2)(k - 6) } $ $q = \dfrac{ 4(k + 2)(k - 6)}{ (k - 10)(k + 2)(k - 6)} $ Notice that $(k - 6)$ and $(k + 2)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 4\cancel{(k + 2)}(k - 6)}{ (k - 10)\cancel{(k + 2)}(k - 6)} $ We are dividing by $k + 2$ , so $k + 2 \neq 0$ Therefore, $k \neq -2$ $q = \dfrac{ 4\cancel{(k + 2)}\cancel{(k - 6)}}{ (k - 10)\cancel{(k + 2)}\cancel{(k - 6)}} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $q = \dfrac{4}{k - 10} ; \space k \neq -2 ; \space k \neq 6 $